Is converting a maximization algorithm into a minimization one a matter of changing max to min?

This might come across as a silly question but I am curious to know if given a maximization algorithm and asked to get the dual (minimization version), it is just a matter of converting all max's into min's and doing other basic adjustments?

If yes, are there any problems where this would not be the case? If not,开发者_StackOverflow中文版 is there a good intuitive reason why this does not work?

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Yes, maximization and minimization problems are basically the same. The solution for max(f(x)) is the same as -min(-f(x)).

When searching game trees this relation is used for example to convert a minimax search into a negamax search. This has the advantage that instead of writing two functions, one for maximizing your score and another for minimizing the opponent's score, you write a single maximizing function but flip the sign of the result of the evaluation function when it's the other person's move.

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It depends on how your maximization algorithm works. Numerical algorithms that need gradients will probably do more than max and min, and other complexities can come up.

However, there is indeed a very easy fix. Maximizing a function f(a,b,c) is equivalent to minimizing -f(a,b,c). So just negate result of the function.

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It works if the problem is obviously symmetrical like finding the maximum vs. a minimum on a 2D-surface. However, since you are quoting the Knapsack problem: that is a whole other class of problems, the optimum cannot be found by applying some max() function in a greedy manner to a vector.